We over a thought experiment of sorts in class today, and I'm trying to look for some help to better wrap my head around it. For this experiment, I ask you to image a frictionless rod rotating horizontal (oriented on the x-axis and spinning "in and out of the page") with a bead threaded around it, its distance from the origin some amount $r_0$ of $r$. It spins at a constant velocity $\omega$. I can be convinced that at $t=0$, the only force on the bead is the rod, parallel to $\hat \theta$. Since it's not really an extension of the rod but threaded around it, it doesn't feel a central force and is just pushed by the rod in a circle.
Now, according to my lecturer, at later $t$ the bead starts to slide off the rod. This is anecdotally plausible, but proving it using mathematics seems to cause me some confusion. If the force vector is always perpendicular to the rod, how could it possibly slide off?
Here's the explanation the lecturer gave:
Use polar coordinates, so the force on an object can be expressed by the following:
$$\vec F = \hat r \vec F_r + \hat \theta F_{\theta}$$
The lecturer then notes $\hat r \vec F_r = 0$. I assume this is because the rod is smooth, so there is no source for a force to occur radially.
$$\vec F = \hat \theta F_{\theta}$$
And, since $\vec F_r = 0$,
$$\hat r a_r = 0$$ $$a_r = \ddot r-r \dot \theta ^2$$
$$ r \dot \theta ^2 = \ddot r$$
Thus, according to the fact that $a_r =0$, if there is no friction on this object it will move outward. Now, other than making deductions using the equations, I have no clue as to why this is true. Sure, according to the equations this makes sense, but intuitively, I have no clue how the bead can mov outward if the force on it is always perpendicular to the rod.
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